Geometric intuition behind Cauchy not having a mean I'm trying to follow the geometric intuition behind the Cauchy distribution in the book "An introduction to probability theory and its applications" by Feller volume 2, first edition. On page 51, he describes the density of the Cauchy distribution as:
$$\gamma_t = \frac{t}{(t^2+x^2)\pi}$$
He then describes an experiment where a ray of light is emitted horizontally onto a vertical mirror from a source O. The light strikes the mirror at point A and the mirror can rotate about a vertical axis passing through A. This is shown in the figure below. The light reflects off the mirror and strikes the wall O is on at a distance $X$ from O.

The first assertion is that if $\phi$ is uniformly distributed between $(-\frac{\pi}{2}, \frac{\pi}{2})$, $X$ has the density given by $\gamma_t$. I managed to prove this. Then he says that its apparent that if this experiment is repeated $n$ times and the average taken, then this average $\frac{X_1+X_2+\dots X_n}{n}$ will have the same distribution as $X_1$. I didn't follow this part. I guess I could derive it from the density, but Feller seems to be hinting at some kind of obvious geometric intuition which I'm completely missing.
 A: I have not read the book, but I think of it in the following way. For any one observation, you expect $x$ to be not that large, but there's clearly a possibility of drawing an angle $\varphi$ such that you get a really large $x$.
If you do this $n$ times and average all the distances $x_1, \dots, x_n$, then do you expect the average to "tighten up" at all? Will it start to hone in around $O$?
The answer is no. You can get angles $\varphi$ such that $x_{n + 1}$ is so extreme that the average jerks around as wildly as $x_1$. Informally I think of this as representing the idea that with the Cauchy distribution, more data doesn't help. There is no learning going on.
I don't know if that's geometric enough for you, but the point is that you are likely enough get distances so large that the average is no more informative than the first observation.
A: I'm adding a partial answer based on some insight I've collected so far in the hope that someone might be able to complete it. Let's simplify the problem and consider proving that $X_1+X_2$ has the same distribution as $2X_1$. $2 X_1$ implies scaling the distribution by $2$. And since $t$ is the scaling parameter of the distribution, it implies placing the mirror at $2t$ instead and repeating the experiment. So, we're basically saying that placing the mirror at $t$ and generating $X_1$ and $X_2$ and adding them is equivalent to placing the mirror at $2t$ and generating $X$. The $X$ has the same distribution as $X_1+X_2$. This becomes equivalent to proving that $\tan(\phi_1)+\tan(\phi_2)$ has the same distribution as $2 \tan(\phi)$ where $\phi_1$, $\phi_2$ and $\phi$ are iid uniform random numbers between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$.
